In first quantization the particle density operator is


In second quantization I have:

$$ n(\vec{x})=\sum_{\alpha,i,j}\langle i|_{\alpha}\delta^{3}(\vec{x}-\vec{x}_{\alpha})|j\rangle_{\alpha}|i\rangle_{\alpha}\langle j|_{\alpha}$$

I know that

$$ \sum_{\alpha}|i\rangle_{\alpha}\langle j|_{\alpha}=a_{i}^{\dagger}a_j$$

How can I introduce the creation and annihilation operators in the first equation?, the sum depend on $\alpha$.

I need to show that



closed as off-topic by Emilio Pisanty, Jon Custer, sammy gerbil, user191954, Kyle Kanos Nov 20 '18 at 10:55

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  • $\begingroup$ You should double-check your second equation. The excerpt $)l\rangle$ should almost certainly read $)|j\rangle$. $\endgroup$ – Emilio Pisanty Nov 18 '18 at 21:22

I don't think your sum on alpha is supposed to collapse into the expression you are trying to show, here is my approach:

A single particle operator such as $n$ may be written in second quantized form according to

$$n = \sum_{\alpha , \beta} \langle \alpha | n | \beta \rangle a_{\alpha}^{\dagger} a_{\beta} $$

for some complete set of states {${\alpha , \beta}$}. Equivalently, we may write this using the field operators

$n = \int dx dy \langle x | n | y \rangle \psi(x)^{\dagger} \psi(y) = \sum_{i} \int dx dy \delta(x-x_i)\langle x | y \rangle \psi(x)^{\dagger} \psi(y) = \sum_{i} \psi(x_i)^{\dagger} \psi(x_i)$

For completeness, I should define the field operator

$$\psi(x) = \sum_i \langle x| i \rangle a_i$$

This may help as well.

  • $\begingroup$ thanks, i am not sure of my expression. I supposed that the second equation depend on alpha becouse I and j are states of the particles and different particles must have different states or not? $\endgroup$ – F.Mark Nov 20 '18 at 0:35

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